Solutions  for  Homework  5    

1. Kittel  2.4  

a)  

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|F 2 |= 1− e[− iM (a⋅Δk )]

1− e[−i(a⋅Δk )]⎛

⎝ ⎜

⎞

⎠ ⎟ 1− e[iM (a⋅Δk )]

1− e[i(a⋅Δk )]⎛

⎝ ⎜

⎞

⎠ ⎟  

 Expanding  the  parentheses  and  using:  

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cos(x) = (eix + e− ix ) /2    

We  get:  

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|F 2 |= 2 − 2cos(Ma⋅ Δk)2 − 2cos(a⋅ Δk)

=sin2(Ma⋅ Δk)sin2(a⋅ Δk)

 

 b)  For  F2  to  be  zero  the  numerator  must  be  zero.    sin[(M(2πh+ε)/2]=0    Using  trig.  Identities:    sin[(M(2πh+ε)/2]=sin(Mπh)cos(Mε/2)+cos(Mπh)sin(Mε/2)=  sin(Mε/2)=0    Therefore  Mε/2=nπ, and for the first zero n=1 and ε=2π/M.

2. Kittel 2.5  

a) The  basis  points  are:  (0,0,0),  a/2(1,1,0).  a/2(1,0,1),  a/2(0,1,1),  a/4(1,1,1),    a/4(-­‐1,1,-­‐1),a/4(-­‐1,-­‐1,1),a/4(1,-­‐1,-­‐1).  

 We  write  G=(n1b1+  n2b2  +  n3b3)    The  structure  factor  is  then  given  by:    

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SG =1+ eiπ (n1 +n2 ) + eiπ (n1 +n3 ) + eiπ (n3 +n2 ) +

eiπ (n1 +n2 +n3 ) / 2 + eiπ (−n1 −n2 +n3 ) / 2 + eiπ (n1 −n2 −n3 ) / 2 + eiπ (−n1 +n2 −n3 ) / 2)  

 b) Let  n=  n1+  n2  +  n3  .  Then  we  can  write  SG  as:  

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SG = (1+ eiπn / 2)(eiπ (n1 +n3 ) + eiπ (n3 +n2 ) + eiπ (n1 +n2 ))    Then  the  zeros  are  n=4m+2  with  m  –integer  coming  from  the  first  factor.  The  second  factor  gives  zeroes  when  one  of  the  n1,  n2  ,  n3  is  odd  AND  the  others  are  even  or  one  of  the  n1,  n2  ,  n3  is  even  AND  the  others  are  odd.  So  for  reflections  with  the  nonzero  structure  factor  1)n  must  be  odd  or  2)n=4m  with  all  even  n1,  n2  ,  n3    

3. Kittel 2.6  

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n(r) =e−2r / a0

πa03  

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The  form  factor  is:  

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f = 4π drn(r) sin(Gr)Gr0

∞

∫ ;  

 Looking  this  up  in  an  integral  table  (say  in  Wikipedia)    

gives:  

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f =16

(4 +G2a02)2  

   Problem  4.    

   

a)  1st:   2acos(30º)=  

           

2nd:  -­‐ acos(30º)+ (2a-­‐asin(30º))=  

-­‐ /2+ 1.5a    

3rd:   2c  

b) =-2 c( cos(30º)+ cos(30º))=-2 c( +0.5 )=- c( +)

=-2 c

=1.5

-3ac =-33/2a2c

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b1 =2π(a2 × a3)a1⋅ (a2 × a3)

=

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b2 =2π(a1 × a3)a1⋅ (a2 × a3)

= =

= =

c) SHKL=

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eigHKL ⋅riunitcell

∑ , where gHKL are reciprocal lattice vectors, ri are coordinates of ith

atom in the unit cell. For graphite we have 4 atoms in the unit cell. The coordinates are:

r1=(0,0,0), r2=(-1/3,2/3,0), r3=(-1/3,2/3,0.5), r4=(-2/3,4/3,0.5) in the coordinates defined by normalized unit vectors.

r1=(0,0,0), r2=(0,a,0), r3=(0,a,c), r4=(0,2a,c) in the coordinates defined by x,y,z. (Note that the top layer is simply shifted by a in the y-direction.

SHKL=1+exp(i((2π/3)H+(4π/3)K))+ exp(i((2π/3)H+(4π/3)K+πL))+ exp(i((4π/3)H+(8π/3)K+πL))=

=1+exp(i((2π/3)H-(2π/3)K))+ exp(i((2π/3)H-(2π/3)K+πL))+ exp(i((-2π/3)H+(2π/3)K+πL))=

= 1+exp(i((2π/3)(H-K))+(-1)L2cos ((2π/3)(H-K)).

or:

SHKL=1+exp(i((2π/3)H+(4π/3)K))+ exp(i((2π/3)H+(4π/3)K+πL))+ exp(i((4π/3)H+(8π/3)K+πL))=

=1+exp(i((2π/3)H+(4π/3)K))+(1+exp(i((2π/3)H+(4π/3)K)))exp exp(i((2π/3)H+(4π/3)K)+πL)=

(1+exp(i((2π/3)H+(4π/3)K)))(1+exp exp(i((2π/3)H+(4π/3)K)+πL))= (1+exp(i((2π/3)(H-K))(1+exp(i((2π/3)(H-K)+πL))=

(1+exp(i((2π/3)(H-K))(1+(-1)Lexp(i((2π/3)(H-K)).

d) S10L=1+exp(i((2π/3)H)+( -1)L2cos ((2π/3)H).